Answer:
[tex]1[/tex]
Step-by-step explanation:
The gist of modular arithmetic in a nutshell: the numbers [tex]a[/tex] and [tex]b[/tex] are considered to be congruent by their modulus [tex]m[/tex] if [tex]m[/tex] is a divisor of their difference.
In mathematics: [tex]a \equiv_{m} b \Leftrightarrow (a - b) \vdots m[/tex]
Exemplifying this: [tex]6 \equiv_{7} -1[/tex] because [tex]6 - (-1) = 6 + 1 = 7[/tex], [tex]7 \vdots 7[/tex].
Let us have following equivalences: [tex]a \equiv_{m} b[/tex] and [tex]c \equiv_{m} d[/tex], then: [tex](a - b) \vdots m[/tex] and [tex](c - d) \vdots m[/tex] by definition.
Properties:
1. [tex]a + c \equiv_{m} b + d \Leftrightarrow ((a + c) - (b + d)) \vdots m \Leftrightarrow (a + c - b - d) \vdots m \Leftrightarrow ((a - b) + (c - d)) \vdots m[/tex].
2. [tex]a - c \equiv_{m} b - d \Leftrightarrow ((a - c) - (b - d)) \vdots m \Leftrightarrow (a - c - b + d) \vdots m \Leftrightarrow ((a - b) - (c - d)) \vdots m[/tex].
3. [tex]ac \equiv_{m} bd \Leftrightarrow (ac - bd) \vdots m \Leftrightarrow (ac - bc - bd + bc) \vdots m \Leftrightarrow (c(a - b) + b(c - d)) \vdots m[/tex].
4. What if we have [tex]a \equiv_{m} b[/tex] twice? If we abide by property 3, we can come to the conclusion that [tex]a^2 \equiv_m b^2[/tex]. It is fair enough that there is room for the equivalence [tex]a^n \equiv_{m} b^n[/tex].
[tex]2001 * 2002 * 2003 + 2004^3 = 2002 * (2001 * 2003) + 2004^3 \equiv_{7} 0 * (2001 * 2003) + 2^3 \equiv_{7} 0 + 8 \equiv_{7} 8 \equiv_{7} 1[/tex]
Notice that [tex]2002 \vdots 7[/tex] already. There is no need to consider the rest multipliers because the product is [tex]0[/tex] anyways, we can cut corners in this regard.
We used property 4.
Recapitulating this: our remainder is [tex]1[/tex].
